A playground is on the flat roof of a city school, h b = 5.40 m above the street

Question

A playground is on the flat roof of a city school, hb = 5.40 m above the street below (see figure). The vertical wall of the building is h = 6.50 m high, to form a 1.1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall.

(a) Find the speed at which the ball was launched.
m/s

(b) Find the vertical distance by which the ball clears the wall.
m

(c) Find the horizontal distance from the wall to the point on the roof where the ball lands.
m

Explanation / Answer

a)

Horizontal velocity, ux= horizontal range/time= 24/2.2= 10.91 m/s

Initial velocity is given by

ux= u cos 53

u= 18.127 m/s

b) Now using 2nd equation of motion

Y= u sin 53*t-0.5*9.8t^2

y= 18.127 sin 53*2.2-0.5*9.8*2.2^2

= 8.133 m

Ball clear the wall by= 8.133-6.5= 1.633 m

c)

Now vertical position of the ball, y= 5.4 m

Again using 2nd equation of motion

5.4= 18.127* sin 53*t-0.5*9.8*t^2

Solving for t

t= 2.517 s

Horizontal distance= 27.458 m

Required distance= 27.458-24= 3.458 m

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Commemt in case any doubt. Good luck

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